Domain & Range: Calculus Help, Graphs

[1calculus1]

Calculus Help, Please. (Domain, Range, Graphs)

Homework Statement

Determine its domain and range.
f(x) = {2x+1, x<0
{2x+2, x => 0

f(x) = {x^2 + 2, x<=1
{2x^2 + 2, x >1

f(x) = { |x| + 1, x<1
{ -x +1, x=>1

Homework Equations

No equations necessary?

The Attempt at a Solution

I'm thinking I could graph it or solve the f(x) by plugging in the x value indicated from the question. I don't really know.


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Homework Statement

Please see the attachment file at the bottom of this post called "Graph".
The question is "Write the function whose graph is given in the figure."

Homework Equations

Slope equation:
( Y2 - Y1 ) / (X2 - X1 )

The Attempt at a Solution

(0 -(-1)) / (-2 -(-1)) = 1
(-1 -1) / (3-(-1)) = -1/2

So now I know the slopes of that might help me with getting the function. However, I'm stuck. Anybody know what to do next?

 

[malawi_glenn]

you don't need to post twice, it is forbidden.

domain is what values of x you can have, and range is what values your function f can "reach".

So start with the first one:
f(x) = {2x+1, x<0
{2x+2, x => 0

What x's can we have? And which f(x) can we get?

 

[1calculus1]

I posted twice due to the fact that I made a mistake on putting this thread here instead of putting it in the "Precalculus" Section.

So, what you're saying is that if I plugged in:
f(x) = 2x + 1, x<0
= 2(-1) + 1 = 1
= 2(-2) + 1 = 3
= 2(-3) + 1 = 5
= 2(-4) + 1 = 7

f(x) = 2x+2, x=> 0
= 2(0) + 2 = 2
= 2(1) + 2 = 4
= 2(2) + 2 = 6
= 2(3) + 2 = 8

So the answer should be domain is X>0 and the range is Y>0 ?
I don't get it.

 

[1calculus1]

Also, the answer from the book is the domain: (-infinity, infinity) and
the range is (-infinity, 1), [2, infinity)

I still don't get it.

 

[||spoon||]

f(x)=2x+1, x<0.

This ststement means that the fuction f(x) is defined (or exists) for all values of x which ate less than zero.

Similarly f(x)=2x+2, x=>0 means that this function is defined for all values of x above and including 0.

If you were to draw these you would have the first function all to the left of x=0 and the second function all above and at x=0.

Since the graphs are straight lines try to think of the shape of the graphs as they go to their limits (most extreme x values) so for the first function negative infinity and zero. Whatever the y values approach, this will give you the range of tge function.

This basically gives you your domain..

 

[||spoon||]

(sorry meant negative infinty and 1)

EDIT: no I didn't lol

 

[1calculus1]

f(x)=2x+1, x<0.

This ststement means that the fuction f(x) is defined (or exists) for all values of x which ate less than zero.

Similarly f(x)=2x+2, x=>0 means that this function is defined for all values of x above and including 0.

If you were to draw these you would have the first function all to the left of x=0 and the second function all above and at x=0.

Since the graphs are straight lines try to think of the shape of the graphs as they go to their limits (most extreme x values) so for the first function negative infinity and zero. Whatever the y values approach, this will give you the range of tge function.

This basically gives you your domain..

But, that doesn't make sense because 2(-1) + 1 = -1
So, shouldn't it be negative infinity and -1 for the first function? Also, I graphed it as you said and this is what I came up with: (Please see the attachment called "Graph2" below this reply).

I'm still confused on how the range is 1 and why the answer in the book is [2, infinity), what's that suppose to mean? Also, just for reference, I'm using "Calculus of A Single Variable -Early Transcendental Functions Third Editions" book.

 

[||spoon||]

Why are you solving f(-1)? You need to solve for f(0) for the first function which gives 1. The other value should be negative infinity, I think you know this. So far the range is (-infinity,1) but this is only half of the question. You still have the second function. Remember this question is about a hybrid function (a function made of other functions).

 

[1calculus1]

Why are you solving f(-1)? You need to solve for f(0) for the first function which gives 1. The other value should be negative infinity, I think you know this. So far the range is (-infinity,1) but this is only half of the question. You still have the second function. Remember this question is about a hybrid function (a function made of other functions).

Now, I feel so dumb but isn't "X < 0" means that the "X" value need to be less than zero? So, less than 0 is -1 therefore the f(X) must be f(-1), right? I don't get why you have to solve for f(0) when it doesn't say X =< 0 to the first function. Or do I always have to assume to put "zero" first in the function and solve it that way THEN get the X value according to the rules indicated (eg. X < 0 )?

Sorry for taking too much of your time. I just really want to get Calculus.

 

[1calculus1]

Hold on. I think I got it.
However, I still don't get my #2 Question:

Homework Statement

Please see the attachment file at the bottom of this post called "Graph".
The question is "Write the function whose graph is given in the figure."

Homework Equations

Slope equation:
( Y2 - Y1 ) / (X2 - X1 )

The Attempt at a Solution

(0 -(-1)) / (-2 -(-1)) = 1
(-1 -1) / (3-(-1)) = -1/2

So now I know the slopes of that might help me with getting the function. However, I'm stuck. Anybody know what to do next?

 

[||spoon||]

no you are right, you need to put a value less than zero. But their are smaller negative numbers which are closer to 0:-0.5,-0.1,-0.000000001. So where do we draw the line? Well we use a number which is infitely small and subtract it from 0, which basically gives zero. We say as x "approaches" zero.

The way this is shown in your answer is by use of different brackets. Note for the first part the range is (-infinity,1) I.e. Using these "(,)" brackets. This means that the value you obtain is infinitely close to the value you put there, or another way of thinking of it is that it will only just not reach this value or will "approach" that value.

Note also inthe second part of the function that the range is given [2,infinity) the square brackets show that the the function actually DOES reach this value.

Also no need to be sorry haha.

 

[1calculus1]

no you are right, you need to put a value less than zero. But their are smaller negative numbers which are closer to 0:-0.5,-0.1,-0.000000001. So where do we draw the line? Well we use a number which is infitely small and subtract it from 0, which basically gives zero. We say as x "approaches" zero.

The way this is shown in your answer is by use of different brackets. Note for the first part the range is (-infinity,1) I.e. Using these "(,)" brackets. This means that the value you obtain is infinitely close to the value you put there, or another way of thinking of it is that it will only just not reach this value or will "approach" that value.

Note also inthe second part of the function that the range is given [2,infinity) the square brackets show that the the function actually DOES reach this value.

Also no need to be sorry haha.

WOW! Thank you so much. Now I know why "[" is used. Plus, I get it! *clap*clap! You're much better than my Calculus teacher. =)

 

[||spoon||]

remember though that with non-linear graphs the range might not be found by the extreme x-values. There may be a minimum/maximum value between the given domain. This is why if you cannot properly visualize the function it is always best to graph it. Even if you think you can visualise it it may still be beneficial to graph the function to be sure. Having it in front of you ofyen makes things easier.

 

[1calculus1]

Oh, I'm also just wondering what does the "U" stands for? (Eg. (-infinity, 0) U (1, infinity)Edit: Thanks for the tip above. I'll keep that in mind.

 

[||spoon||]

the "U" symbol represents the expression "in union with" but I suppose you can just say another very technical term... "and"

in this context it basically means the set of values from negative infinity to 0 "and" from 1 to infinity.

 

[||spoon||]

Now for your second question!

I think you have already realized this is a hybrid function with two parts. You have also correctly found the gradient (slope) of both of these parts. Good stuff!

Now, as you can see the parts are linear functions and you know (i assume) that the general form of a linear function is of the type y=mx+c, where m is the gradient and c is the y-intercept.

What you need to do is sub in the information you are given and tgen limit the domain for each part so it fits the function we want...

So, for the first part let's use the point (-2,0) and the gradient you found to be 1.
Given y=mx+c it follows that:
0=(1)(-2)+c
therefore c=2.

Now we arrive at y=x+2. However this is still not correct because y=x+2 has a domain of all real numbers whereas in the graph you provided tgis part of the hybrid function clearly does not. We are going to have to "restrict" the domain.

You can see from your graph the values which y=x+2 does exist, this is from (-2,-1) so we say that this part of the hybrid function follows the equation:

y=x+2, -2<x<-1

Note that I have used < not =< because the points you gave were in () brackets, if they were in [] brackets you would need to use <=.

Now try the second part and when youbare done write the final hybrid function as those in your first question.

 

[1calculus1]

Now for your second question!

I think you have already realized this is a hybrid function with two parts. You have also correctly found the gradient (slope) of both of these parts. Good stuff!

Now, as you can see the parts are linear functions and you know (i assume) that the general form of a linear function is of the type y=mx+c, where m is the gradient and c is the y-intercept.

What you need to do is sub in the information you are given and tgen limit the domain for each part so it fits the function we want...

So, for the first part let's use the point (-2,0) and the gradient you found to be 1.
Given y=mx+c it follows that:
0=(1)(-2)+c
therefore c=2.

Now we arrive at y=x+2. However this is still not correct because y=x+2 has a domain of all real numbers whereas in the graph you provided tgis part of the hybrid function clearly does not. We are going to have to "restrict" the domain.

You can see from your graph the values which y=x+2 does exist, this is from (-2,-1) so we say that this part of the hybrid function follows the equation:

y=x+2, -2<x<-1

Note that I have used < not =< because the points you gave were in () brackets, if they were in [] brackets you would need to use <=.

Now try the second part and when youbare done write the final hybrid function as those in your first question.

Thanks for the reply. Now, I follow all except how you have gotten the y-value of -1. Isn't suppose to be 0 since that -2 + 2 = 0? Also, the book's answer is -2<=x<=-1. Do you think they might have made a mistake? Because my teacher said that there are few mistakes in the book.

I'll try the second part now from what you've told me:

y=mx+c
(3, -1)
-1 = -1/2(3) + c
c = 1/2

y=-1/2x+1/2
and then I'm stuck again because I have no idea how to get the restriction.

 

[1calculus1]

Hold on, now that I think about it, do I just have to look at the graph to determine the restriction? Since the graph (referring to the first function) does not extend to the x-value (domain) of -2 and -1, does that mean that that's the answer? It's between those points? But, I still don't get it since the book says -2<=x<=-1. How can I determine when to put the "equal" thing in the restriction?

Also, it's kind of the same thing with the second function since the graph of x-values does not extend to 3 and -1 does that mean that -1<x<3? But, the book says that the answer is -1<x<=3. I simply don't get the fact when to put the "equal" sign in the restriction area. Please help.I feel so dumb. =(

 

[||spoon||]

im not sure what you mean about -2+2=0... How are you getting this?

If the books answer uses "<=" etc. then the graph either uses [] brackets for the endpoints or the lines have filled in black dots at the endpoints. If there are filled in black dots this means the same as square brackets, that the equation DOES exist for the said values. The alternative is hollow dots "o" at the end of the line segments, this is equivalent to () brackets, meaning that the equation APPROACHES the given value.

Perhaps your book has this but you did not add the detail to your graph?

Your equation y=-1/2x+1/2 is correct. Now domain is what x-valures the equation is defined (exists) over. So if you look at the graph. What are it's most extreme x-values?

 

[1calculus1]

im not sure what you mean about -2+2=0... How are you getting this?

If the books answer uses "<=" etc. then the graph either uses [] brackets for the endpoints or the lines have filled in black dots at the endpoints. If there are filled in black dots this means the same as square brackets, that the equation DOES exist for the said values. The alternative is hollow dots "o" at the end of the line segments, this is equivalent to () brackets, meaning that the equation APPROACHES the given value.

Perhaps your book has this but you did not add the detail to your graph?

Your equation y=-1/2x+1/2 is correct. Now domain is what x-valures the equation is defined (exists) over. So if you look at the graph. What are it's most extreme x-values?

For the -2+2=0.. You know how the equation is y=x+2, I just assumed that since we have the x value from the first function of -2, we could just plug it in there to find the y-value. Therefore making it (-2, 0). But, obviously, it's wrong because the x values are in between the -2 and -1.

Yes, you're right. I definitely did not put the filled black dots on the graph. All of the points should have had filled black dots on there. However, if that is the case, I'm still confused due to the fact that the answer to the second function is -1<x<=3. How come there's a < instead of it being <= like everything else? Because as you have said, if the filled black dots are there, everything should have that equal sign in the restriction.

So what went wrong?

 

[1calculus1]

Or, from the -1<x<=3, the -1 is "used" or in between the two functions so does that mean that it's different therefore making it not totally equal?

 

[||spoon||]

when you subbed in -2 for x you are finding the y value of the equation when x=-2, which is 0. But remember that domain is the possible X-Values... No need to find the corresponding y-values.

I think that if there is a black dot at (-1,1) then it should be "<= =>" if it has open it should be "<>", if the book does not specify I would've assumed that wether it be "<>" or. "<= =>" then both sides of this point would be the same. I.e both approaching or both equal to -1.

I wouldn't stress about it ;D

 

[1calculus1]

when you subbed in -2 for x you are finding the y value of the equation when x=-2, which is 0. But remember that domain is the possible X-Values... No need to find the corresponding y-values.

I think that if there is a black dot at (-1,1) then it should be "<= =>" if it has open it should be "<>", if the book does not specify I would've assumed that wether it be "<>" or. "<= =>" then both sides of this point would be the same. I.e both approaching or both equal to -1.

I wouldn't stress about it ;D

Alright. So any tips on getting the restrictions? Should I just look at the graph on where the points extend and what not?
Thanks for still being here answering my questions. =)

 

[||spoon||]

ok just think... What would the graph look like without the restrictions?? Now how do i make it look like it does?

Basically (in this case) you do just need to look at the graph and note which values OF X does each part exist over? Do not confuse domain and range.

I'm glad to have helped, stick in with calculus! It is one of the most useful and interesting parts of mathematics =D

 

[1calculus1]

ok just think... What would the graph look like without the restrictions?? Now how do i make it look like it does?

Basically (in this case) you do just need to look at the graph and note which values OF X does each part exist over? Do not confuse domain and range.

I'm glad to have helped, stick in with calculus! It is one of the most useful and interesting parts of mathematics =D

I thought so. Thanks for all the help again. You have no idea how much you've helped me. =)